## NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.1

Ex 2.1 Class 9 Maths Question 1.

Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.

(i) 4×2 – 3x + 7

(ii) y2 + √2

(iii) 3 √t + t√2

(iv) y+ (frac { 2 }{ y })

(v) x10+ y3+t50

Solution:

(i) We have 4×2 – 3x + 7 = 4×2 – 3x + 7×0

It is a polynomial in one variable i.e., x

because each exponent of x is a whole number.

(ii) We have y2 + √2 = y2 + √2y0

It is a polynomial in one variable i.e., y

because each exponent of y is a whole number.

(iii) We have 3 √t + t√2 = 3 √t1/2 + √2.t

It is not a polynomial, because one of the exponents of t is (frac { 1 }{ 2 }),

which is not a whole number.

(iv) We have y + (y+frac { 2 }{ y }) = y + 2.y-1

It is not a polynomial, because one of the exponents of y is -1,

which is not a whole number.

(v) We have x10+ y3 + t50

Here, exponent of every variable is a whole number, but x10 + y3 + t50 is a polynomial in x, y and t, i.e., in three variables.

So, it is not a polynomial in one variable.

Ex 2.1 Class 9 Maths Question 2.

Write the coefficients of x2 in each of the following

(i) 2 + x2 + x

(ii) 2 – x2 + x3

(iii) (frac { pi }{ 2 }) x2 + x

(iv) √2 x – 1

Solution:

(i) The given polynomial is 2 + x2 + x.

The coefficient of x2 is 1.

(ii) The given polynomial is 2 – x2 + x3.

The coefficient of x2 is -1.

(iii) The given polynomial is (frac { pi }{ 2 } { x }^{ 2 }) + x.

The coefficient of x2 is (frac { pi }{ 2 }).

(iv) The given polynomial is √2 x – 1.

The coefficient of x2 is 0.

Ex 2.1 Class 9 Maths Question 3.

Give one example each of a binomial of degree 35, and of a monomial of degree 100.

Solution:

(i) Abmomial of degree 35 can be 3×35 -4.

(ii) A monomial of degree 100 can be √2y100.

Ex 2.1 Class 9 Maths Question 4.

Write the degree of each of the following polynomials.

(i) 5×3+4×2 + 7x

(ii) 4 – y2

(iii) 5t – √7

(iv) 3

Solution:

(i) The given polynomial is 5×3 + 4×2 + 7x.

The highest power of the variable x is 3.

So, the degree of the polynomial is 3.

(ii) The given polynomial is 4- y2. The highest

power of the variable y is 2.

So, the degree of the polynomial is 2.

(iii) The given polynomial is 5t – √7 . The highest power of variable t is 1. So, the degree of the polynomial is 1.

(iv) Since, 3 = 3x° [∵ x°=1]

So, the degree of the polynomial is 0.

Ex 2.1 Class 9 Maths Question 5.

Classify the following as linear, quadratic and cubic polynomials.

(i) x2+ x

(ii) x – x3

(iii) y + y2+4

(iv) 1 + x

(v) 3t

(vi) r2

(vii) 7×3

Solution:

(i) The degree of x2 + x is 2. So, it is a quadratic polynomial.

(ii) The degree of x – x3 is 3. So, it is a cubic polynomial.

(iii) The degree of y + y2 + 4 is 2. So, it is a quadratic polynomial.

(iv) The degree of 1 + x is 1. So, it is a linear polynomial.

(v) The degree of 3t is 1. So, it is a linear polynomial.

(vi) The degree of r2 is 2. So, it is a quadratic polynomial.

(vii) The degree of 7×3 is 3. So, it is a cubic polynomial.

### NCERT Solutions for Class 9 Maths Chapter 2 Polynomials (बहुपद) (Hindi Medium) Ex 2.1

### NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2

Question 1.

Find the value of the polynomial 5x – 4×2 + 3 at

(i) x = 0

(ii) x = – 1

(iii) x = 2

Solution:

1et p(x) = 5x – 4×2 + 3

(i) p(0) = 5(0) – 4(0)2 + 3 = 0 – 0 + 3 = 3

Thus, the value of 5x – 4×2 + 3 at x = 0 is 3.

(ii) p(-1) = 5(-1) – 4(-1)2 + 3

= – 5x – 4×2 + 3 = -9 + 3 = -6

Thus, the value of 5x – 4×2 + 3 at x = -1 is -6.

(iii) p(2) = 5(2) – 4(2)2 + 3 = 10 – 4(4) + 3

= 10 – 16 + 3 = -3

Thus, the value of 5x – 4×2 + 3 at x = 2 is – 3.

Question 2.

Find p (0), p (1) and p (2) for each of the following polynomials.

(i) p(y) = y2 – y +1

(ii) p (t) = 2 +1 + 2t2 -t3

(iii) P (x) = x3

(iv) p (x) = (x-1) (x+1)

Solution:

(i) Given that p(y) = y2 – y + 1.

∴ P(0) = (0)2 – 0 + 1 = 0 – 0 + 1 = 1

p(1) = (1)2 – 1 + 1 = 1 – 1 + 1 = 1

p(2) = (2)2 – 2 + 1 = 4 – 2 + 1 = 3

(ii) Given that p(t) = 2 + t + 2t2 – t3

∴p(0) = 2 + 0 + 2(0)2 – (0)3

= 2 + 0 + 0 – 0=2

P(1) = 2 + 1 + 2(1)2 – (1)3

= 2 + 1 + 2 – 1 = 4

p( 2) = 2 + 2 + 2(2)2 – (2)3

= 2 + 2 + 8 – 8 = 4

(iii) Given that p(x) = x3

∴ p(0) = (0)3 = 0, p(1) = (1)3 = 1

p(2) = (2)3 = 8

(iv) Given that p(x) = (x – 1)(x + 1)

∴ p(0) = (0 – 1)(0 + 1) = (-1)(1) = -1

p(1) = (1 – 1)(1 +1) = (0)(2) = 0

P(2) = (2 – 1)(2 + 1) = (1)(3) = 3

Question 3.

Verify whether the following are zeroes of the polynomial, indicated against them.

(i) p(x) = 3x + 1,x = –(frac { 1 }{ 3 })

(ii) p (x) = 5x – π, x = (frac { 4 }{ 5 })

(iii) p (x) = x2 – 1, x = x – 1

(iv) p (x) = (x + 1) (x – 2), x = – 1,2

(v) p (x) = x2, x = 0

(vi) p (x) = 1x + m, x = – (frac { m }{ 1 })

(vii) P (x) = 3×2 – 1, x = – (frac { 1 }{ sqrt { 3 } }),(frac { 2 }{ sqrt { 3 } })

(viii) p (x) = 2x + 1, x = (frac { 1 }{ 2 })

Solution:

(i) We have , p(x) = 3x + 1

(ii) We have, p(x) = 5x – π

∴ (p(-frac { 1 }{ 3 } )quad =quad 3(-frac { 1 }{ 3 } )quad +quad 1quad =quad -1quad +quad 1quad =quad 0)

(iii) We have, p(x) = x2 – 1

∴ p(1) = (1)2 – 1 = 1 – 1=0

Since, p(1) = 0, so x = 1 is a zero of x2 -1.

Also, p(-1) = (-1)2 -1 = 1 – 1 = 0

Since p(-1) = 0, so, x = -1, is also a zero of x2 – 1.

(iv) We have, p(x) = (x + 1)(x – 2)

∴ p(-1) = (-1 +1) (-1 – 2) = (0)(- 3) = 0

Since, p(-1) = 0, so, x = -1 is a zero of (x + 1)(x – 2).

Also, p( 2) = (2 + 1)(2 – 2) = (3)(0) = 0

Since, p(2) = 0, so, x = 2 is also a zero of (x + 1)(x – 2).

(v) We have, p(x) = x2

∴ p(o) = (0)2 = 0

Since, p(0) = 0, so, x = 0 is a zero of x2.

(vi) We have, p(x) = lx + m

(vii) We have, p(x) = 3×2 – 1

(viii) We have, p(x) = 2x + 1

∴ (p(frac { 1 }{ 2 } )quad =quad 2(frac { 1 }{ 2 } )+1=quad 1+1quad =quad 2)

Since, (p(frac { 1 }{ 2 } )) ≠ 0, so, x = (frac { 1 }{ 2 }) is not a zero of 2x + 1.

Question 4.

Find the zero of the polynomial in each of the following cases

(i) p(x)=x+5

(ii) p (x) = x – 5

(iii) p (x) = 2x + 5

(iv) p (x) = 3x – 2

(v) p (x) = 3x

(vi) p (x)= ax, a≠0

(vii) p (x) = cx + d, c ≠ 0 where c and d are real numbers.

Solution:

(i) We have, p(x) = x + 5. Since, p(x) = 0

⇒ x + 5 = 0

⇒ x = -5.

Thus, zero of x + 5 is -5.

(ii) We have, p(x) = x – 5.

Since, p(x) = 0 ⇒ x – 5 = 0 ⇒ x = -5

Thus, zero of x – 5 is 5.

(iii) We have, p(x) = 2x + 5. Since, p(x) = 0

⇒ 2x + 5 =0

⇒ 2x = -5

⇒ x = (frac { -5 }{ 2 })

Thus, zero of 2x + 5 is (frac { -5 }{ 2 }) .

(iv) We have, p(x) = 3x – 2. Since, p(x) = 0

⇒ 3x – 2 = 0

⇒ 3x = 2

⇒ x = (frac { 2 }{ 3 })

Thus, zero of 3x – 2 is (frac { 2 }{ 3 })

(v) We have, p(x) = 3x. Since, p(x) = 0

⇒ 3x = 0 ⇒ x = 0

Thus, zero of 3x is 0.

(vi) We have, p(x) = ax, a ≠ 0.

Since, p(x) = 0 => ax = 0 => x-0

Thus, zero of ax is 0.

(vii) We have, p(x) = cx + d. Since, p(x) = 0

⇒ cx + d = 0 ⇒ cx = -d ⇒ ( x =-frac { d }{ c })

Thus, zero of cx + d is (-frac { d }{ c })

### NCERT So1utions for C1ass 9 Maths Chapter 2 Polynomials Ex 2.3

Question 1.

Find the remainder when x3 + 3×2 + 3x + 1 is divided by

(i) x + 1

(ii) x – (frac { 1 }{ 2 })

(iii) x

(iv) x + π

(v) 5 + 2x

Solution:

Let p(x) = x3 + 3×2 + 3x +1

(i) The zero of x + 1 is -1.

∴ p(-1) = (-1)3 + 3(-1)2 + 3(-1) +1

= -1 + 3- 3 + 1 = 0

Thus, the required remainder = 0

(ii) The zero of (x-frac { 1 }{ 2 }) is (frac { 1 }{ 2 })

Thus, the required remainder = (frac { 27 }{ 8 })

(iii) The zero of x is 0.

∴ p(0) = (0)3 + 3(0)2 + 3(0) + 1

= 0 + 0 + 0 + 1 = 1

Thus, the required remainder = 1.

(iv) The zero of x + π is -π.

p(-π) = (-π)3 + 3(- π)22 + 3(- π) +1

= -π3 + 3π2 + (-3π) + 1

= – π3 + 3π2 – 3π +1

Thus, the required remainder is -π3 + 3π2 – 3π+1.

(v) The zero of 5 + 2x is (-frac { 5 }{ 2 }) .

Thus, the required remainder is (-frac { 27 }{ 8 }) .

Question 2.

Find the remainder when x3 – ax2 + 6x – a is divided by x – a.

Solution:

We have, p(x) = x3 – ax2 + 6x – a and zero of x – a is a.

∴ p(a) = (a)3 – a(a)2 + 6(a) – a

= a3 – a3 + 6a – a = 5a

Thus, the required remainder is 5a.

Question 3.

Check whether 7 + 3x is a factor of 3×3+7x.

Solution:

We have, p(x) = 3×3+7x. and zero of 7 + 3x is (-frac { 7 }{ 3 }).

Since,( (-frac { 490 }{ 9 })) ≠ 0

i.e. the remainder is not 0.

∴ 3×3 + 7x is not divisib1e by 7 + 3x.

Thus, 7 + 3x is not a factor of 3×3 + 7x.

### NCERT So1utions for Class 9 Maths Chapter 2 Polynomials Ex 2.4

Question 1.

Determine which of the following polynomials has (x +1) a factor.

(i) x3+x2+x +1

(ii) x4 + x3 + x2 + x + 1

(iii) x4 + 3×3 + 3×2 + x + 1

(iv) x3 – x2 – (2 +√2 )x + √2

Solution:

The zero of x + 1 is -1.

(i) Let p (x) = x3 + x2 + x + 1

∴ p (-1) = (-1)3 + (-1)2 + (-1) + 1 .

= -1 + 1 – 1 + 1

⇒ p (- 1) = 0

So, (x+ 1) is a factor of x3 + x2 + x + 1.

(ii) Let p (x) = x4 + x3 + x2 + x + 1

∴ P(-1) = (-1)4 + (-1)3 + (-1)2 + (-1)+1

= 1 – 1 + 1 – 1 + 1

⇒ P (-1) ≠ 1

So, (x + 1) is not a factor of x4 + x3 + x2 + x+ 1.

(iii) Let p (x) = x4 + 3×3 + 3×2 + x + 1 .

∴ p (-1)= (-1)4 + 3 (-1)3 + 3 (-1)2 + (- 1) + 1

= 1 – 3 + 3 – 1 + 1 = 1

⇒ p (-1) ≠ 0

So, (x + 1) is not a factor of x4 + 3×3 + 3×2 + x+ 1.

(iv) Let p (x) = x3 – x2 – (2 + √2) x + √2

∴ p (- 1) =(- 1)3- (-1)2 – (2 + √2)(-1) + √2

= -1 – 1 + 2 + √2 + √2

= 2√2

⇒ p (-1) ≠ 0

So, (x + 1) is not a factor of x3 – x2 – (2 + √2) x + √2.

Question 2.

Use the Factor Theorem to determine whether g (x) is a factor of p (x) in each of the following cases

(i) p (x)= 2×3 + x2 – 2x – 1, g (x) = x + 1

(ii) p(x)= x3 + 3×2 + 3x + 1, g (x) = x + 2

(iii) p (x) = x3 – 4×2 + x + 6, g (x) = x – 3

Solution:

(i) We have, p (x)= 2×3 + x2 – 2x – 1 and g (x) = x + 1

∴ p(-1) = 2(-1)3 + (-1)2 – 2(-1) – 1

= 2(-1) + 1 + 2 – 1

= -2 + 1 + 2 -1 = 0

⇒ p(-1) = 0, so g(x) is a factor of p(x).

(ii) We have, p(x) x3 + 3×2 + 3x + 1 and g(x) = x + 2

∴ p(-2) = (-2)3 + 3(-2)2+ 3(-2) + 1

= -8 + 12 – 6 + 1

= -14 + 13

= -1

⇒ p(-2) ≠ 0, so g(x) is not a factor of p(x).

(iii) We have, = x3 – 4×2 + x + 6 and g (x) = x – 3

∴ p(3) = (3)3 – 4(3)2 + 3 + 6

= 27 – 4(9) + 3 + 6

= 27 – 36 + 3 + 6 = 0

⇒ p(3) = 0, so g(x) is a factor of p(x).

Question 3.

Find the value of k, if x – 1 is a factor of p (x) in each of the following cases

(i) p (x) = x2 + x + k

(ii) p (x) = 2×2 + kx + √2

(iii) p (x) = kx2 – √2 x + 1

(iv) p (x) = kx2 – 3x + k

Solution:

For (x – 1) to be a factor of p(x), p(1) should be equal to 0.

(i) Here, p(x) = x2 + x + k

Since, p(1) = (1)2 +1 + k

⇒ p(1) = k + 2 = 0

⇒ k = -2.

(ii) Here, p (x) = 2×2 + kx + √2

Since, p(1) = 2(1)2 + k(1) + √2

= 2 + k + √2 =0

k = -2 – √2 = -(2 + √2)

(iii) Here, p (x) = kx2 – √2 x + 1

Since, p(1) = k(1)2 – (1) + 1

= k – √2 + 1 = 0

⇒ k = √2 -1

(iv) Here, p(x) = kx2 – 3x + k

p(1) = k(1)2 – 3(1) + k

= k – 3 + k

= 2k – 3 = 0

⇒ k = (frac { 3 }{ 4 })

Question 4.

Factorise

(i) 12×2 – 7x +1

(ii) 2×2 + 7x + 3

(iii) 6×2 + 5x – 6

(iv) 3×2 – x – 4

Solution:

(i) We have,

12×2 – 7x + 1 = 12×2 – 4x- 3x + 1

= 4x (3x – 1 ) -1 (3x – 1)

= (3x -1) (4x -1)

Thus, 12×2 -7x + 3 = (2x – 1) (x + 3)

(ii) We have, 2×2 + 7x + 3 = 2×2 + x + 6x + 3

= x(2x + 1) + 3(2x + 1)

= (2x + 1)(x + 3)

Thus, 2×2 + 7x + 3 = (2x + 1)(x + 3)

(iii) We have, 6×2 + 5x – 6 = 6×2 + 9x – 4x – 6

= 3x(2x + 3) – 2(2x + 3)

= (2x + 3)(3x – 2)

Thus, 6×2 + 5x – 6 = (2x + 3)(3x – 2)

(iv) We have, 3×2 – x – 4 = 3×2 – 4x + 3x – 4

= x(3x – 4) + 1(3x – 4) = (3x – 4)(x + 1)

Thus, 3×2 – x – 4 = (3x – 4)(x + 1)

Question 5.

Factorise

(i) x3 – 2×2 – x + 2

(ii) x3 – 3×2 – 9x – 5

(iii) x3 + 13×2 + 32x + 20

(iv) 2y3 + y2 – 2y – 1

Solution:

(i) We have, x3 – 2×2 – x + 2

Rearranging the terms, we have x3 – x – 2×2 + 2

= x(x2 – 1) – 2(x2 -1) = (x2 – 1)(x – 2)

= [(x)2 – (1)2](x – 2)

= (x – 1)(x + 1)(x – 2)

[∵ (a2 – b2) = (a + b)(a-b)]

Thus, x3 – 2×2 – x + 2 = (x – 1)(x + 1)(x – 2)

(ii) We have, x3 – 3×2 – 9x – 5

= x3 + x2 – 4×2 – 4x – 5x – 5 ,

= x2 (x + 1) – 4x(x + 1) – 5(x + 1)

= (x + 1)(x2 – 4x – 5)

= (x + 1)(x2 – 5x + x – 5)

= (x + 1)[x(x – 5) + 1(x – 5)]

= (x + 1)(x – 5)(x + 1)

Thus, x3 – 3×2 – 9x – 5 = (x + 1)(x – 5)(x +1)

(iii) We have, x3 + 13×2 + 32x + 20

= x3 + x2 + 12×2 + 12x + 20x + 20

= x2(x + 1) + 12x(x +1) + 20(x + 1)

= (x + 1)(x2 + 12x + 20)

= (x + 1)(x2 + 2x + 10x + 20)

= (x + 1)[x(x + 2) + 10(x + 2)]

= (x + 1)(x + 2)(x + 10)

Thus, x3 + 13×2 + 32x + 20

= (x + 1)(x + 2)(x + 10)

(iv) We have, 2y3 + y2 – 2y – 1

= 2y3 – 2y2 + 3y2 – 3y + y – 1

= 2y2(y – 1) + 3y(y – 1) + 1(y – 1)

= (y – 1)(2y2 + 3y + 1)

= (y – 1)(2y2 + 2y + y + 1)

= (y – 1)[2y(y + 1) + 1(y + 1)]

= (y – 1)(y + 1)(2y + 1)

Thus, 2y3 + y2 – 2y – 1

= (y – 1)(y + 1)(2y +1)

### NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5

Question 1.

Use suitable identities to find the following products

(i) (x + 4)(x + 10)

(ii) (x+8) (x -10)

(iii) (3x + 4) (3x – 5)

(iv) (y2+ (frac { 3 }{ 2 })) (y2– (frac { 3 }{ 2 }))

(v) (3 – 2x) (3 + 2x)

Solution:

(i) We have, (x+ 4) (x + 10)

Using identity,

(x+ a) (x+ b) = x2 + (a + b) x+ ab.

We have, (x + 4) (x + 10) = x2+(4 + 10) x + (4 x 10)

= x2 + 14x+40

(ii) We have, (x+ 8) (x -10)

Using identity,

(x + a) (x + b) = x2 + (a + b) x + ab

We have, (x + 8) (x – 10) = x2 + [8 + (-10)] x + (8) (- 10)

= x2 – 2x – 80

(iii) We have, (3x + 4) (3x – 5)

Using identity,

(x + a) (x + b) = x2 + (a + b) x + ab

We have, (3x + 4) (3x – 5) = (3x)2 + (4 – 5) x + (4) (- 5)

= 9×2 – x – 20

Question 2.

Evaluate the following products without multiplying directly

(i) 103 x 107

(ii) 95 x 96

(iii) 104 x 96

Solution:

(i)We have, 103 x 107 = (100 + 3) (100 + 7)

= ( 100)2 + (3 + 7) (100)+ (3 x 7)

[Using (x + a)(x + b) = x2 + (a + b)x + ab]

= 10000 + (10) x 100 + 21

= 10000 + 1000 + 21=11021

(ii) We have, 95 x 96 = (100 – 5) (100 – 4)

= ( 100)2 + [(- 5) + (- 4)] 100 + (- 5 x – 4)

[Using (x + a)(x + b) = x2 + (a + b)x + ab]

= 10000 + (-9) + 20 = 9120

= 10000 + (-900) + 20 = 9120

(iii) We have 104 x 96 = (100 + 4) (100 – 4)

= (100)2-42

[Using (a + b)(a -b) = a2– b2]

= 10000 – 16 = 9984

Question 3.

Factorise the following using appropriate identities

(i) 9×2 + 6xy + y2

(ii) 4y2-4y + 1

(iii) x2 – (frac { { y }^{ 2 } }{ 100 })

Solution:

(i) We have, 9×2 + 6xy + y2

= (3x)2 + 2(3x)(y) + (y)2

= (3x + y)2

[Using a2 + 2ab + b2 = (a + b)2]

= (3x + y)(3x + y)

(ii) We have, 4y2 – 4y + 12

= (2y)2 + 2(2y)(1) + (1)2

= (2y -1)2

[Using a2 – 2ab + b2 = (a- b)2]

= (2y – 1)(2y – 1 )

Question 4.

Expand each of the following, using suitable identity

(i) (x+2y+ 4z)2

(ii) (2x – y + z)2

(iii) (- 2x + 3y + 2z)2

(iv) (3a -7b – c)z

(v) (- 2x + 5y – 3z)2

(vi) [ (frac { 1 }{ 4 })a –(frac { 1 }{ 4 })b + 1] 2

Solution:

We know that

(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx

(i) (x + 2y + 4z)2

= x2 + (2y)2 + (4z)2 + 2 (x) (2y) + 2 (2y) (4z) + 2(4z) (x)

= x2 + 4y2 + 16z2 + 4xy + 16yz + 8 zx

(ii) (2x – y + z)2 = (2x)2 + (- y)2 + z2 + 2 (2x) (- y)+ 2 (- y) (z) + 2 (z) (2x)

= 4×2 + y2 + z2 – 4xy – 2yz + 4zx

(iii) (- 2x + 3y + 2z)2 = (- 2x)2 + (3y)2 + (2z)2 + 2 (- 2x) (3y)+ 2 (3y) (2z) + 2 (2z) (- 2x)

= 4×2 + 9y2 + 4z2 – 12xy + 12yz – 8zx

(iv) (3a -7b- c)2 = (3a)2 + (- 7b)2 + (- c)2 + 2 (3a) (- 7b) + 2 (- 7b) (- c) + 2 (- c) (3a)

= 9a2 + 49b2 + c2 – 42ab + 14bc – 6ac

(v)(- 2x + 5y- 3z)2 = (- 2x)2 + (5y)2 + (- 3z)2 + 2 (- 2x) (5y) + 2 (5y) (- 3z) + 2 (- 3z) (- 2x)

= 4×2 + 25y2 + 9z2 – 20xy – 30yz + 12zx

Question 5.

Factorise

(i) 4 x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz

(ii) 2×2 + y2 + 8z2 – 2√2xy + 4√2yz – 8xz

Solution:

(i) 4×2 + 9y2 + 16z2 + 12xy – 24yz – 16xz

= (2x)2 + (3y)2 + (- 4z)2 + 2 (2x) (3y) + 2 (3y) (- 4z) + 2 (- 4z) (2x)

= (2x + 3y – 4z)2 = (2x + 3y + 4z) (2x + 3y – 4z)

(ii) 2×2 + y2 + 8z2 – 2√2xy + 4√2yz – 8xz

= (- √2x)2 + (y)2 + (2 √2z)2y + 2(- √2x) (y)+ 2 (y) (2√2z) + 2 (2√2z) (- √2x)

= (- √2x + y + 2 √2z)2

= (- √2x + y + 2 √2z) (- √2x + y + 2 √2z)

Question 6.

Write the following cubes in expanded form

Solution:

We have, (x + y)3 = x3 + y3 + 3xy(x + y) …(1)

and (x – y)3 = x3 – y3 – 3xy(x – y) …(2)

(i) (2x + 1)3 = (2x)3 + (1)3 + 3(2x)(1)(2x + 1) [By (1)]

= 8×3 + 1 + 6x(2x + 1)

= 8×3 + 12×2 + 6x + 1

(ii) (2a – 3b)3 = (2a)3 – (3b)3 – 3(2a)(3b)(2a – 3b) [By (2)]

= 8a3 – 27b3 – 18ab(2a – 3b)

= 8a3 – 27b3 – 36a2b + 54ab2

Question 7.

Evaluate the following using suitable identities

(i) (99)3

(ii) (102)3

(iii) (998)3

Solution:

(i) We have, 99 = (100 -1)

∴ 993 = (100 – 1)3

= (100)3 – 13 – 3(100)(1)(100 -1)

[Using (a – b)3 = a3 – b3 – 3ab (a – b)]

= 1000000 – 1 – 300(100 – 1)

= 1000000 -1 – 30000 + 300

= 1000300 – 30001 = 970299

(ii) We have, 102 =100 + 2

∴ 1023 = (100 + 2)3

= (100)3 + (2)3 + 3(100)(2)(100 + 2)

[Using (a + b)3 = a3 + b3 + 3ab (a + b)]

= 1000000 + 8 + 600(100 + 2)

= 1000000 + 8 + 60000 + 1200 = 1061208

(iii) We have, 998 = 1000 – 2

∴ (998)3 = (1000-2)3

= (1000)3– (2)3 – 3(1000)(2)(1000 – 2)

[Using (a – b)3 = a3 – b3 – 3ab (a – b)]

= 1000000000 – 8 – 6000(1000 – 2)

= 1000000000 – 8 – 6000000 +12000

= 994011992

Question 8.

Factorise each of the following

(i) 8a3 +b3 + 12a2b+6ab2

(ii) 8a3 -b3-12a2b+6ab2

(iii) 27-125a3 -135a+225a2

(iv) 64a3 -27b3 -144a2b + 108ab2

Solution:

(i) 8a3 +b3 +12a2b+6ab2

= (2a)3 + (b)3 + 6ab(2a + b)

= (2a)3 + (b)3 + 3(2a)(b)(2a + b)

= (2 a + b)3

[Using a3 + b3 + 3 ab(a + b) = (a + b)3]

= (2a + b)(2a + b)(2a + b)

(ii) 8a3 – b3 – 12o2b + 6ab2

= (2a)3 – (b)3 – 3(2a)(b)(2a – b)

= (2a – b)3

[Using a3 + b3 + 3 ab(a + b) = (a + b)3]

= (2a – b) (2a – b) (2a – b)

(iii) 27 – 125a3 – 135a + 225a2

= (3)3 – (5a)3 – 3(3)(5a)(3 – 5a)

= (3 – 5a)3

[Using a3 + b3 + 3 ab(a + b) = (a + b)3]

= (3 – 5a) (3 – 5a) (3 – 5a)

(iv) 64a3 -27b3 -144a2b + 108ab2

= (4a)3 – (3b)3 – 3(4a)(3b)(4a – 3b)

= (4a – 3b)3

[Using a3 – b3 – 3 ab(a – b) = (a – b)3]

= (4a – 3b)(4a – 3b)(4a – 3b)

Question 9.

Verify

(i) x3 + y3 = (x + y)-(x2 – xy + y2)

(ii) x3 – y3 = (x – y) (x2 + xy + y2)

Solution:

(i) ∵ (x + y)3 = x3 + y3 + 3xy(x + y)

⇒ (x + y)3 – 3(x + y)(xy) = x3 + y3

⇒ (x + y)[(x + y)2-3xy] = x3 + y3

⇒ (x + y)(x2 + y2 – xy) = x3 + y3

Hence, verified.

(ii) ∵ (x – y)3 = x3 – y3 – 3xy(x – y)

⇒ (x – y)3 + 3xy(x – y) = x3 – y3

⇒ (x – y)[(x – y)2 + 3xy)] = x3 – y3

⇒ (x – y)(x2 + y2 + xy) = x3 – y3

Hence, verified.

Question 10.

Factorise each of the following

(i) 27y3 + 125z3

(ii) 64m3 – 343n3

[Hint See question 9]

Solution:

(i) We know that

x3 + y3 = (x + y)(x2 – xy + y2)

We have, 27y3 + 125z3 = (3y)3 + (5z)3

= (3y + 5z)[(3y)2 – (3y)(5z) + (5z)2]

= (3y + 5z)(9y2 – 15yz + 25z2)

(ii) We know that

x3 – y3 = (x – y)(x2 + xy + y2)

We have, 64m3 – 343n3 = (4m)3 – (7n)3

= (4m – 7n)[(4m)2 + (4m)(7n) + (7n)2]

= (4m – 7n)(16m2 + 28mn + 49n2)

Question 11.

Factorise 27×3 +y3 +z3 -9xyz.

Solution:

We have,

27×3 + y3 + z3 – 9xyz = (3x)3 + (y)3 + (z)3 – 3(3x)(y)(z)

Using the identity,

x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx)

We have, (3x)3 + (y)3 + (z)3 – 3(3x)(y)(z)

= (3x + y + z)[(3x)3 + y3 + z3 – (3x × y) – (y × 2) – (z × 3x)]

= (3x + y + z)(9×2 + y2 + z2 – 3xy – yz – 3zx)

Question 12.

Verify that

x3 +y3 +z3 – 3xyz = (frac { 1 }{ 2 }) (x + y+z)[(x-y)2 + (y – z)2 +(z – x)2]

Solution:

R.H.S

= (frac { 1 }{ 2 })(x + y + z)[(x – y)2+(y – z)2+(z – x)2]

= (frac { 1 }{ 2 }) (x + y + 2)[(x2 + y2 – 2xy) + (y2 + z2 – 2yz) + (z2 + x2 – 2zx)]

= (frac { 1 }{ 2 }) (x + y + 2)(x2 + y2 + y2 + z2 + z2 + x2 – 2xy – 2yz – 2zx)

= (frac { 1 }{ 2 }) (x + y + z)[2(x2 + y2 + z2 – xy – yz – zx)]

= 2 x (frac { 1 }{ 2 }) x (x + y + z)(x2 + y2 + z2 – xy – yz – zx)

= (x + y + z)(x2 + y2 + z2 – xy – yz – zx)

= x3 + y3 + z3 – 3xyz = L.H.S.

Hence, verified.

Question 13.

If x + y + z = 0, show that x3 + y3 + z3 = 3 xyz.

Solution:

Since, x + y + z = 0

⇒ x + y = -z (x + y)3 = (-z)3

⇒ x3 + y3 + 3xy(x + y) = -z3

⇒ x3 + y3 + 3xy(-z) = -z3 [∵ x + y = -z]

⇒ x3 + y3 – 3xyz = -z3

⇒ x3 + y3 + z3 = 3xyz

Hence, if x + y + z = 0, then

x3 + y3 + z3 = 3xyz

Question 14.

Without actually calculating the cubes, find the value of each of the following

(i) (- 12)3 + (7)3 + (5)3

(ii) (28)3 + (- 15)3 + (- 13)3

Solution:

(i) We have, (-12)3 + (7)3 + (5)3

Let x = -12, y = 7 and z = 5.

Then, x + y + z = -12 + 7 + 5 = 0

We know that if x + y + z = 0, then, x3 + y3 + z3 = 3xyz

∴ (-12)3 + (7)3 + (5)3 = 3[(-12)(7)(5)]

= 3[-420] = -1260

(ii) We have, (28)3 + (-15)3 + (-13)3

Let x = 28, y = -15 and z = -13.

Then, x + y + z = 28 – 15 – 13 = 0

We know that if x + y + z = 0, then x3 + y3 + z3 = 3xyz

∴ (28)3 + (-15)3 + (-13)3 = 3(28)(-15)(-13)

= 3(5460) = 16380

Question 15.

Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given

(i) Area 25a2 – 35a + 12

(ii) Area 35y2 + 13y – 12

Solution:

Area of a rectangle = (Length) x (Breadth)

(i) 25a2 – 35a + 12 = 25a2 – 20a – 15a + 12 = 5a(5a – 4) – 3(5a – 4) = (5a – 4)(5a – 3)

Thus, the possible length and breadth are (5a – 3) and (5a – 4).

(ii) 35y2+ 13y -12 = 35y2 + 28y – 15y -12

= 7y(5y + 4) – 3(5y + 4) = (5 y + 4)(7y – 3)

Thus, the possible length and breadth are (7y – 3) and (5y + 4).

Question 16.

What are the possible expressions for the dimensions of the cuboids whose volumes are given below?

(i) Volume 3×2 – 12x

(ii) Volume 12ky2 + 8ky – 20k

Solution:

Volume of a cuboid = (Length) x (Breadth) x (Height)

(i) We have, 3×2 – 12x = 3(x2 – 4x)

= 3 x x x (x – 4)

∴ The possible dimensions of the cuboid are 3, x and (x – 4).

(ii) We have, 12ky2 + 8ky – 20k

= 4[3ky2 + 2ky – 5k] = 4[k(3y2 + 2y – 5)]

= 4 x k x (3y2 + 2y – 5)

= 4k[3y2 – 3y + 5y – 5]

= 4k[3y(y – 1) + 5(y – 1)]

= 4k[(3y + 5) x (y – 1)]

= 4k x (3y + 5) x (y – 1)

Thus, the possible dimensions of the cuboid are 4k, (3y + 5) and (y -1).

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